# Electrical – Relation between settling time and natural frequency for a critically damped system

control system

In an article by Wie et al. (1989), for the purpose of calculating the gains of a controller, the equations of motion are approximated by the expression for a damped harmonic oscillation:

$$\ddot{\theta} + d \dot{\theta} + \frac{k}{2} \theta = 0$$

where, of course:

$$d = 2 \zeta \omega _n \quad ; \quad \frac{k}{2} = \omega^2 _n$$

In the design example presented in the article, a settling time of $$T_s = 50 \text{ sec}$$
is assumed. Now, the article states that for a critically damped response, $$\omega _n = 0.158 \text{ rad/s.}$$

Indeed, using this natural frequency to calculate the controller gains yields responses that settle after about 50 seconds: However, when using the relation between the settling time and the natural frequency for a critically damped system as presented in this answer, one obtaines:

$$\omega_n \approx \frac{5.8335}{T_s} = \frac{5.8335}{50} = 0.117 \text{ rad/s.}$$

So the question is, how was the natural frequency calculated in the article?

You need to be careful when using settle time approximations that relate $$\\omega_n\$$ and $$\\zeta\$$ to a $$\T_s\$$.
First, $$\T_s\$$ is defined as the time where the signal remains within $$\\pm x\$$% of the final value. Typically the settle limits are $$\\pm 2\$$% or $$\\pm 5\$$%, and these limits impact all second order system settle time approximations. Yes, every 2nd order settle time equation is an approximation, and they all differ as a function of $$\\zeta\$$.
Next, $$\omega_n \approx \frac{5.8335}{T_s},$$ is only a numerical approximation to the normalized unit step response for a critically damped system (i.e., $$\\zeta=1\$$), which is $$\theta(t)= \left(1-(1+\omega_n t)e^{-\omega_n t}\right)u(t),$$ where $$\u(t)\$$ is a unit step (aka Heaviside step). The settle time approximation for this is eloquently derived in your "this answer" link using a $$\\pm 2\$$% window to numerically arrive at $$\\omega_n \approx \frac{5.8335}{T_s}\$$.
When I look at the graphs of the data you shared, it looks like the 2% settle time for the critically damped trace is closer to 35. This results in $$\\omega_n\approx\frac{5.8335}{35}= 0.167\$$ rad/sec. A result is closer to the actual $$\\omega_n\$$ in the article. Better graphs, and clarification on which trace is actually the critically damped one, will improve this estimate.
Sorry, but I do not have free access to your article, so cannot answer your question directly, but most likely the authors knew $$\\omega_n\$$ directly.